10d + 25q = 130

Divide through by 5

 2d + 5q = 26

2 is the least coefficient in absolute value,
and 26 is already a multiple of 2, so we write
5 in terms of its nearest multiple of 2

  2d + (4+1)q = 26
  2d + 4d + q = 26
 d + 2q + q/2 = 13

Isolate the fraction:

          q/2 = 13 - d - 2q

The right side is an integer and the left side
is non-negative, so both sides equal some non-
negative integer A.

q/2 = A  and 13 - d - 2q = A

q = 2A, substituting in

   13 - d - 2q = A
13 - d - 2(2A) = A
   13 - d - 4A = A
       13 - 5A = d

The minimum number of dimes is 0 
and the maximum number of dimes is 13,

so   0 ≦ d ≦ 13

  0 ≦ 13 - 5A ≦ 13
  -13 ≦ -5A ≦ 0
    2.6 ≧ A ≧ 0

The minimum number of quarters is 0 
and the maximum number of quarters is 5,

so   0 ≦ q ≦ 5

    0 ≦ 2A ≦ 5
    0 ≦ A ≦ 2.5
    
So A = 0, A = 1 or A = 2


A    13-5A =  d     2A = q
0  13-5(0) = 13   2(0) = 0
1  13-5(1) =  8   2(1) = 2
2  13-5(2) =  3   2(2) = 4


So

13 dimes and no quarters.
 8 dimes and  2 quarters.
 3 dimes and  4 quarters.

Edwin